Chapter 16: Aqueous Ionic

Equilibrium

Mrs. Brayfield

16.2: Buffers

Most solutions will rapidly change pH upon the addition

of an acid or base

Those that do not are called a buffer

A buffer is a solution that resists pH changes by neutralizing

added acid or added base

Buffers contain significant amounts of both a WEAK acid

and its conjugate base

Or a weak base and its conjugate acid

Blood is a very common buffer

It resists pH changes very well or else you would die easily

Buffers

The reason why you

must have a weak

acid/base and its

conjugate is because

the weak acid/base

does not dissociate

much

Buffers

Why can’t we use strong acids/bases in buffers?

It takes much more base to change the pH of a weak acid

solution because there is a large reservoir of undissociated

weak acid

Calculate the concentration of the starting acid for HCl and

HC2H3O2 (Ka = 1.8x10-5)when the pH of both solutions is 2.00

Common Ion Effect

The common ion effect is the tendency for a common ion

to decrease the ionization of a weak acid or weak base

Calculating the pH of a Buffer

Calculate the pH of a buffer solution that is 0.200M

HC2H3O2 and 0.100M NaC2H3O2

𝐾𝑎 =𝐶2𝐻3𝑂2

− [𝐻+]

[𝐻𝐶2𝐻3𝑂2]= 1.8 × 10−5 =

(.1 + 𝑥)(𝑥)

.2 − 𝑥

[HC2H3O2] [H+] [C2H3O2-]

I 0.200 0 0.100

C – x +x +x

E .2 – x x .1 + x

pH of Buffers

To find the pH of buffers, we just rearrange Ka:

𝐾𝑎 =𝐻3𝑂

+ [𝐴−]

[𝐻𝐴]

𝐻3𝑂+ = 𝐾𝑎

[𝐻𝐴]

[𝐴−]

We can now take the log of both sides to find the pH:

− log 𝐻3𝑂+ = −𝑙𝑜𝑔𝐾𝑎 − 𝑙𝑜𝑔

[𝐻𝐴]

[𝐴−]

𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔[𝐴−]

[𝐻𝐴]

𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔[𝑏𝑎𝑠𝑒]

[𝑎𝑐𝑖𝑑]

pH of Buffers

𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔[𝑏𝑎𝑠𝑒]

[𝑎𝑐𝑖𝑑]

This is called the Henderson-Hasselbalch equation and

we can easily calculate the pH of a buffer solution from

initial concentrations, as long as x is small (we can make

our assumption)

pH of Buffers Example

Calculate the pH of a buffer solution that is 0.250M HCN

and 0.170M KCN. For HCN Ka = 4.9 x 10-10 (pKa = 9.31).

Henderson-Hasselbalch Equation

So, when can we use this equation?

Basically whenever we can approximate

You should look for the initial concentrations to be at least 102

to 103 LARGER than the Ka

Also, since the log is a ratio of concentrations [HA]/[A–],

amounts of acid/base in moles can be substituted for

concentration (since volumes cancel – see page 606)

Base Buffers

Same calculations as with our acid buffers:

𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔[𝑏𝑎𝑠𝑒]

[𝑎𝑐𝑖𝑑]

Just make sure that you change Kb to Ka

𝐾𝑤 = 𝐾𝑎 × 𝐾𝑏

OR

14 = 𝑝𝐾𝑎 + 𝑝𝐾𝑏

Base Buffers Example

Calculate the pH of a solution containing 0.60M NH3 and

0.30M NH4Cl (Kb = 4.75).

16.3: Buffer Effectiveness

For a buffer to be effective it must neutralize moderate

amounts of added acid or base

There are two factors that influence the effectiveness of a

buffer:

1. Relative amounts of the acid and conjugate base

2. Absolute concentrations of the acid and conjugate base

Buffer Effectiveness

We do not need to calculate the change in pH of a buffer

BUT we can see on the example on page 610 that buffers

with equal amounts of acid and conjugate base is more

resistant to pH change

Therefore it is a more effective buffer

As a guideline an effective buffer MUST have a

[base]/[acid] ratio in the range of 0.10 to 10

Basically the two concentrations should not differ by more

than a factor or 10 to be effective

Buffer Capacity

Buffer capacity is the amount of acid or base that can be

added to a buffer without destroying its effectiveness

Buffer capacity increases with increasing concentrations

of the buffer components

There are more species to deal with changes in pH

Buffer capacity also increases as the concentrations of the

buffer components get closer to each other

The overall capacity of the buffer increases

Last Note on Buffers

Without being given any data on the initial concentrations

we can make some inferences about those values:

If pH < pKa, the acid has a higher concentration than the base

If pH > pKa, the base has a higher concentration than the acid

Homework Problems: #1, 2, 4, 8, 10, 12, 14, 16, 20, 28, 30,

32

16.4: Titrations and pH Curves

An acid-base titration is where a basic (or acidic) solution

of unknown concentration is reacted with an acidic (or

basic) solution of known concentration

The known solution is slowly added to the unknown solution

while pH is monitored

We can monitor pH using indicators, or a substance whose color

depends on the pH

As the acid and base react, they neutralize each other

This is called the equivalence point (when the number of moles of

base EQUAL the number of moles of acid)

Titrations

This is called a pH curve:

Titrations

Buret

Strong Acid and Strong Base Titration

Calculations

Find the pH after adding different amounts of 0.100M NaOH to 25mL of 0.100M HCl:

Initial pH

pH after adding:

5.00mL NaOH

10.0mL NaOH

15.0mL NaOH

20.0mL NaOH

25.0mL NaOH

30.0mL NaOH

35.0mL NaOH

40.0mL NaOH

50.0mL NaOH

SA and SB Titration Calculations

Titration Calculation Example 1

Calculate the pH in the titration in the previous example

after adding a total of 60.0mL of 0.200M HNO3 (50.0mL

of 0.200M NaOH titrated with 0.200M HNO3)

Weak Acid with Strong Base Titration

Calculations

Find the pH after adding different amounts of 0.100M NaOH to 25mL of 0.100M HCHO2:

Initial pH

pH after adding:

5.00mL NaOH

10.0mL NaOH

15.0mL NaOH

20.0mL NaOH

25.0mL NaOH

30.0mL NaOH

35.0mL NaOH

40.0mL NaOH

50.0mL NaOH

WA and SB Titration Calculations

Titration Calculation Example 2

Determine the pH at the equivalence point for the

titration between HNO2 and KOH in the previous

example (40mL 0.10M HNO2 titrated with 0.20M KOH)

Indicators

Titrations can be monitored with either a pH meter or

an indicator

With an indicator we rely on when that indicator turns

colors, otherwise known as the endpoint, to determine

the equivalence point

Indicators

Many Different Indicators…

Homework Problems: #36, 38, 40, 43, 46, 50, 52, 54

16.5: Solubility Equilibria and Ksp

Remember the solubility rules from a long time ago…

All sodium, potassium, ammonium, and nitrate salts are soluble

in water

AP does not require you to know the rest of the rules

Using equilibrium, we now can better understand the

solubility of an ionic compound:

𝐶𝑎𝐹2 𝑠 ↔ 𝐶𝑎2+ 𝑎𝑞 + 2𝐹− 𝑎𝑞

𝐾𝑠𝑝 = 𝐶𝑎2+ [𝐹−]2

Where Ksp is the solubility product constant

Remember that solids are omitted from the equilibrium

expression

Ksp

Ksp is a measure

of the solubility

of a compound

See Appendix IIC

for a complete

table of values

Molar Solubility

Molar solubility (S) is the solubility in units of moles per

liter (mol/L), which can be calculated from Ksp

For example:

𝐴𝑔𝐶𝑙 𝑠 ↔ 𝐴𝑔+ 𝑎𝑞 + 𝐶𝑙− 𝑎𝑞

Where S is the concentration of AgCl that dissolves

(molar solubility)

[Ag+] [Cl–]

I 0.00 0.00

C + S + S

E S S

𝐾𝑠𝑝 = 𝐴𝑔+ [𝐶𝑙−]

𝐾𝑠𝑝 = 𝑆2

𝑆 = 𝐾𝑠𝑝 = 1.77 × 10−10

𝑆 = 1.33 × 10−5𝑀

So we can say that the molar solubility of AgCl is 1.33 x 10-5M

[Ag+] [Cl–]

I 0.00 0.00

C + S + S

E S S

Molar Solubility Example 1

Calculate the molar solubility of Fe(OH)2 in pure water

Molar Solubility Example 2

The molar solubility of AgBr in pure water is 7.3 x 10-7M.

Calculate Ksp.

Common Ion Effect

When a solution already has one of the compounds ions

in solution, the solubility of that compound is affected

We simply use Le Châtelier's principle to explain the

equilibrium shift

The solubility of any compound is lower in solution containing

a common ion that in pure water

Common Ion Effect Example

Calculate the molar solubility of CaF2 in a solution

containing 0.250M Ca(NO3)2.

Effect of pH on Solubility

Treat this just like a common ion effect problem:

The solubility of an ionic compound with a strongly basic

or weakly basic anion increases with increasing acidity

(decreasing pH)

For example OH–, S2–, CO32–

16.6: Precipitation

We know that precipitation can occur when two

solutions are mixed and produces an insoluble compound

BUT even compounds that are “insoluble” dissolve, if only

slightly, in water

Going back to Q…

We can look at Q and compare it to Ksp to determine if

any more solid will dissolve

Q

Remember from chapter 14:

If Q < Ksp, the solution is unsaturated and more solid can

dissolve

If Q = Ksp, the solution is saturated and no more solid will

dissolve

If Q > Ksp, the solution is supersaturated and solid will

precipitate out

Q Example

If we have solutions of 0.0600M Pb(NO3)2 and 0.0158M

NaBr, will a precipitate form in the mixed solution?

Homework Problems: #56, 58, 60, 62, 70, 72

16.7: Complex Ion Equilibria

A complex ion contains a central metal ion that is bound

to one or more ligands

A ligand is a neutral molecule or ion that acts as a Lewis

base with the central metal ion

For the complex ion in Ag(H2O)2+, water is the ligand

Why learn about complex ions?

If we have a barely soluble compound, we can introduce a

ligand to the solution to get the compound to dissolve more

(remove one product to shift equilibrium – Le Châtelier's)

Review problems: #80, 82, 86, 90

http://www.youtube.com/watch?v=8Fdt5WnYn1k&list=PL

8dPuuaLjXtPHzzYuWy6fYEaX9mQQ8oGr&index=32